Re: [xsl] concat all items in a sequence

Subject: Re: [xsl] concat all items in a sequence
From: David Carlisle <davidc@xxxxxxxxx>
Date: Wed, 31 Aug 2005 14:03:21 +0100
  <xsl:value-of select="count(.|/root/node[contains(text,
  'def')]/preceding-sibling::node)"/>

  This complains because one <node> contains two <text> elements,


<xsl:value-of select="1+count(/root/node[text[contains(.,'def')]]/preceding-sibling::node)"/>



  Regarding true()=elem/contains(...) it's also really good using the
  slash operator :) but I can't see how to squeeze it in here... (why
  the true()= part david?)


You wouldn't need to squeeze it, it would just fit, but the version
michael (I think) suggested using text[contains(. is better.
However,
elem/contains(...)  is a sequence of booleans and you want to know
if any of them is true. as always if you want to know if any item
in a sequence is equal to something you canm se = and it does an
implicit quantification over the sequence.


David

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